Equilíbrio Iônico Aquoso

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Chapter 17 Aqueous Ionic Equilibrium 369 then do an equilibrium calculation: [CH₃NH₃⁺], [H₃O⁺] pH set up ICE table pH = -log [H₃O⁺] Solution: mol base = mol acid = mol = 0.004375 mol. Then total volume = L + L HBr = + = 0.0542 then [CH₃NH₃⁺] = 0.004375 mol = 0.0807196 Set up an ICE table: + = CH₃NH₂(aq) + H₃O⁺(aq) [CH₃NH₃⁺] [CH₃NH₂] [H₃O⁺] Initial 0.0807196 =0.00 Change -x +x +x Equil 0.0807196 x +x +x = [CH₃NH₃⁺] = 2.2727 10⁻¹¹ = 0.0807196 Assume that x is small (x 0.0807), 0.0807196 x = 2.2727 X 10⁻¹¹ = 0.0807196 and x = 1.3544 X 10⁻⁶ = [H₃O⁺]. Confirm that the assumption is valid. 1.3544 0.0807106 X 100% = 0.0017%